The curve C has parametric equations $x = 2 \, ext{cos} \, t, \, y = \, ext{sqrt{3}} \, ext{cos} \, 2t, \, 0 \leq t \leq \pi$ (a) Find an expression for \( \frac{dy}{dx} \) in terms of \( t \) - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 1

To find the normal line at point P where ( t = \frac{2\pi}{3} ):

1. Substitute ( t = \frac{2\pi}{3} ) into the parametric equations: [ x = 2 \cos \left( \frac{2\pi}{3} \right) = -1, \quad y = \sqrt{3} \cos \left( \frac{4\pi}{3} \right) = -\frac{\sqrt{3}}{2} ]

2. Calculate ( \frac{dy}{dx} ) at ( t = \frac{2\pi}{3} ): Using the previously found ( \frac{dy}{dx} ), we find the slope of the tangent line at P: [ \frac{dy}{dx} = 2\sqrt{3} \cos \left( \frac{2\pi}{3} \right) = -\sqrt{3} ]

3. The gradient of the normal line is the negative reciprocal: [ m = \frac{1}{\sqrt{3}} ]

4. The equation of the normal: [ y - \left( -\frac{\sqrt{3}}{2} \right) = \frac{1}{\sqrt{3}} (x + 1) ]

5. Rearranging gives: [ \sqrt{3}y + x - 1 = 0 ] Multiplying through by 2 leads to the desired form: [ 2x - 2\sqrt{3}y - 1 = 0. ]

To find the coordinates of point Q where line l intersects curve C again:

1. Substitute ( y ) from line l into the parametric equation for curve C: [ y = \frac{2}{\sqrt{3}}x + \frac{1}{\sqrt{3}}\text{, where }\ 2x - 2\sqrt{3}y - 1 = 0\text{.} ]

2. Replace ( y ) in the parametric equation: [ \sqrt{3} \cos 2t = \frac{2}{\sqrt{3}}(2\cos t) + \frac{1}{\sqrt{3}} ]

3. Simplify and use identities:

   • Equate and solve for ( t ) using trigonometric identities. Find roots leading back to angle forms.

4. Once found, substitute back: [ x = 2 \cos t, \quad y = \sqrt{3} \cos 2t ext{ to get } Q.
   ] Substitute the found ( t ) value to get the exact coordinates: [ Q = \left( \frac{5}{3}, \frac{7}{18}\sqrt{3} \right) ]